(a) The ball's height <em>y</em> at time <em>t</em> is given by
<em>y</em> = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve <em>y</em> = 0 for <em>t</em> :
0 = (20 m/s) sin(40º) <em>t</em> - 1/2 <em>g t</em> ²
0 = <em>t</em> ((20 m/s) sin(40º) - 1/2 <em>g t</em> )
<em>t</em> = 0 or (20 m/s) sin(40º) - 1/2 <em>g t</em> = 0
The first time refers to where the ball is initially launched, so we omit that solution.
(20 m/s) sin(40º) = 1/2 <em>g t</em>
<em>t</em> = (40 m/s) sin(40º) / <em>g</em>
<em>t</em> ≈ 2.6 s
(b) At its maximum height, the ball has zero vertical velocity. In the vertical direction, the ball is in free fall and only subject to the downward acceleration <em>g</em>. So
0² - ((20 m/s) sin(40º))² = 2 (-<em>g</em>) <em>y</em>
where <em>y</em> in this equation refers to the maximum height of the ball. Solve for <em>y</em> :
<em>y</em> = ((20 m/s) sin(40º))² / (2<em>g</em>)
<em>y</em> ≈ 8.4 m
The final velocity is 2.7 m/s
Explanation:
We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.
Therefore we can write:
where:
is the mass of the putty
is the initial velocity of the putty (we take its direction as positive direction)
is the mass of the ball
is the initial velocity of the ball (at rest)
is the final combined velocity of the two putty+ball
Re-arranging the equation and substituting the values, we find the final combined velocity:
And the positive sign indicates their final direction is the same as the initial direction of the putty.
Learn more about momentum here:
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Answer:
Acceleration
Explanation:
Acceleration is the rate of change in velocity
speed is how fast an object is moving
Velocity is how fast an object is moving in the particular direction
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